Решите задачу:
ОДЗ x-3>0⇒x>3,x+3>0⇒x>-3,(x+3)/(x-3)>0⇒ + _ + --------(-3)------(3)--------- x<-3 U X.3<br>x∈(3;∞) Перейдем к основанию 2 -log(2)(x-3)+log(2)(x+3)-1/log(2)[(x+3)/(x-3)]>0 log(2)[(x+3)/(x-3)] -1/log(2)[(x+3)/(x-3)]>0 [log²(2)[(x+3)/(x-3)]-1]/log(2)[(x+3)/(x-3)]>0 (log(2)[(x+3)/(x-3)]-1)(log(2)[(x+3)/(x-3)]+1)/log(2)[(x+3)/(x-3)]>0 log(2)[(x+3)/(x-3)]=a (a-1)(a+1)/a>0 a=1 a=-1 a=0 _ + _ + ---------(-1)----------(0)-----------(1)----------- -11 1){log(2)[(x+3)/(x-3)]>-1⇒(x+3)/(x-3)>1/2 (1) {log(2)[(x+3)/(x-3)]<0⇒(x+3)(x-3)<1 (2)<br>(1) (x+3)/(x-3)-1/2>0 (2x+6-x+3)/(x-3)>0 (x+9)/(x-3)>0 x=-9 x=3 x<-9 U x>3 (2) (x+3)/(x-3)-1<0<br>(x+3-x+3)/(x-3)<0<br>6/(x-3)<0<br>x-3<0<br>x<3<br>нет решения 2)log(2)[(x+3)/(x-3)]>1 (x+3)/(x-3)>2 (x+3)/(x-3)-2>0 (x+3-2x+6)/(x-3)>0 (9-x)/(x-3)>0 (x-9)/(x-3)<0<br>x=9 x=3 3Ответ x∈(3;9)